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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A 1200 turn coil of mean area $500 \;cm^2 $ is held perpendicular to a uniform field of $4 \times 10^{-4}$ tesla. The resistance of coil is $20 \Omega$ . What will be the average induced current when coil turns through $180^{\circ}$ in $0.1\;$second?

$(a)\;2.4\;mA \\(b)\;0.24\;mA \\ (c)\;24\;mA \\ (d)\;120\;mA $

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1 Answer

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Final position after the coil has turned $180^{\circ}$ is
Initially $\theta=0$ and final $\theta= 180^{\circ}$
$\phi_1=NBA$
$\phi_2=-NBA$
$e_{av}=\large\frac{2NBA}{\Delta t}=\frac{2 \times 1200 \times 4 \times 10^{-4} \times 500 \times 10^{-4}}{0.1}$
$\qquad=0.48\;V$
$I =\large\frac{E_{aV}}{R} =\frac{0.48}{20}$
$\qquad= 24 \times 10^{-3}A$
Hence c is the correct answer.
answered Mar 25, 2014 by meena.p
 

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